Question

Bài 1 : A= (\(\dfrac{3-3\sqrt{x}}{x-9}-1\)) : ( \(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\))

a, Rút gọn

b,tìm x để A < 1

c,Tìm x thuộc Z để A thuộc Z

Bài 2 : A =\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

a, Rút gọn

b, Tính A khi x = 5- \(2\sqrt{6}\)

c, Tìm A min

d, Tìm x để A= 1/2

e, C/m :A < 2/3

Answer 1

Bài 2: 

a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)

d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)

\(\Leftrightarrow11\sqrt{x}=1\)

hay x=1/121